Course 1.2Lesson 2 of 3

Rationalising the denominator

Rationalising removes the square root from a denominator by multiplying the top and bottom by the same number, which does not change the value. For one square root, one over root a becomes root a over a. For a denominator such as a plus root b, multiply by the conjugate a minus root b, and the product is a squared minus b, with no root. The conjugate of root a minus root b is root a plus root b. We can check the answer with decimal values, and rationalising helps to simplify sums and to find values with a calculator.

about 25 min1 simulation2 practice questions

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Why rationalise?

The number $\dfrac{1}{\sqrt{2}}$ is hard to work with, because we divide by an irrational number. It is easier to have a whole number in the denominator. The process of removing the square root from the denominator is called rationalising.

There is also a practical reason. Before calculators, to find $\dfrac{1}{\sqrt{2}}$ you had to divide by $1.41421\ldots$ , which is a long job. But $\dfrac{\sqrt{2}}{2}$ needs only a division by $2$ . Today it still helps, because it is easier to compare numbers and to add fractions when the denominators are whole numbers.

The value does not change. We multiply the top and the bottom by the same number, which is the same as multiplying by $1$ .

One square root

\displaystyle \frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a}

Example: $\dfrac{1}{\sqrt{7}} = \dfrac{\sqrt{7}}{7}$ .

Another example: $\dfrac{5}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}$ .

A third: $\dfrac{6}{2\sqrt{3}} = \dfrac{6\sqrt{3}}{2 \times 3} = \sqrt{3}$ .

A sum or difference with a square root

If the denominator is $a + \sqrt{b}$ , multiply by $a - \sqrt{b}$ . This is the conjugate. The product of a number and its conjugate has no square root, by the identity $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ .

\displaystyle \frac{1}{a + \sqrt{b}} = \frac{1}{a + \sqrt{b}} \times \frac{a - \sqrt{b}}{a - \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}

Example 1. Rationalise $\dfrac{1}{2 + \sqrt{3}}$ .

\displaystyle \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}

Example 2. Rationalise $\dfrac{1}{\sqrt{5} - \sqrt{2}}$ . The conjugate is $\sqrt{5} + \sqrt{2}$ .

\displaystyle \frac{1}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} + \sqrt{2}}{5 - 2} = \frac{\sqrt{5} + \sqrt{2}}{3}

Example 3. A number on top. Rationalise $\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$ . The conjugate of the bottom is $\sqrt{3} + 1$ .

\displaystyle \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}

Example 4. A sum of two fractions. Simplify $\dfrac{1}{2 + \sqrt{3}} + \dfrac{1}{2 - \sqrt{3}}$ . Rationalise each: $\dfrac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$ (Example 1), and in the same way $\dfrac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}$ . The sum is $(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$ .

Change $a$ and $b$ . The two bars never part. Multiplying the top and the bottom by the same number changes how a number looks, not its value.

Checking the answer

We can check with a calculator. $\sqrt{5} \approx 2.236$ and $\sqrt{2} \approx 1.414$ . Then $\dfrac{1}{2.236 - 1.414} = \dfrac{1}{0.822} \approx 1.217$ . And $\dfrac{2.236 + 1.414}{3} = \dfrac{3.650}{3} \approx 1.217$ . They agree.

Practise with new numbers

Try it in Python

This program rationalises $\dfrac{1}{a + \sqrt{b}}$ for four pairs, and it compares the two values.

Common mistakes

Multiplying only the top or only the bottom. Both must be multiplied by the same number.
Using the wrong conjugate. For $a + \sqrt{b}$ , the conjugate is $a - \sqrt{b}$ , and not $-a + \sqrt{b}$ .
Squaring wrongly. $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ , with a minus sign.
Leaving the root in the bottom of the answer. Check that the denominator is a whole number.

Remember these

To rationalise $\dfrac{1}{\sqrt{a}}$ , multiply by $\dfrac{\sqrt{a}}{\sqrt{a}}$ . The result is $\dfrac{\sqrt{a}}{a}$ .
For $\dfrac{1}{a + \sqrt{b}}$ , multiply by the conjugate $a - \sqrt{b}$ . The result is $\dfrac{a - \sqrt{b}}{a^2 - b}$ .
The value does not change. We only multiply by $1$ .